Asked by Kayla

A 62kg cyclist changes the speed of a 12kg bicycle from 8.2m/s to 12.7m/s. Determine the work done.

I've already answered & gotten this question correct, however it made me confused about the concept.

I found Ek. & Ek to find change in Ek which was 3479.85. He did work against both friction & inertia & yet this one formula covered both & I didn't have to add anything together. Why not? As far as I understood, to get Work total you need to do the formula for each work (friction & inertia) then add.

Thank you!! ☺
Answered by bobpursley
No, the work done was the difference in kinetic energies, assuming the work to overcome friction at each speed was constant.
Answered by Kayla
So does that mean if I have work against friction & acceleration I can use that one formula to cover both? :)
Answered by bobpursley
Yes, if the friction involved in attaining different speeds was the same....in real life, it is not. Friction of air is dependent on velocity squared as a rule of thumb.
Answered by Kayla
Ok, that's super helpful thank you very much!! :)
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